1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Marianna [84]
3 years ago
6

A man pulls a sled at a constant velocity across a horizontal snow surface. if a force of 80 n is being applied to the sled rope

at an angle of 53° to the ground, what is the force of friction between sled and snow?
Physics
2 answers:
algol [13]3 years ago
4 0

Answer:

The force of friction between sled and snow is 48.14 N

Explanation:

It is given that,

Force applied on the sled, F = 80 N

The angle with the sled and the ground is 53 degrees. We need to find the force of friction between sled and snow.

The horizontal force acting in the horizontal direction is given by :

F_x=F\ cos\theta

F_x=80\times cos(53)

F_x=48.14\ N

So, the force of friction between sled and snow is 48.14 N. Hence, this is the required solution.

balandron [24]3 years ago
3 0
A constant velocity implies the two forces must be equal and opposite.
Friction acts horizontal to the ground, therefore we must find the force applied to the sled rope that acts horizontal to the ground.
Do this by resolving:
Force = 80cos53
The force opposing this is equal, and so also = 80cos53 = 48 N (2 sig. fig.)
You might be interested in
Which of the following is a theory stating that one plate is forced beneath another plate?
morpeh [17]

Answer:

theroy of plate tectonics

7 0
3 years ago
Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0
ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

F_c=m\frac{v^2}{r}  (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

K=\frac{1}{2}mv^2

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}

you replace this value of v in the equation (1). Also, you replace the values of r and m:

F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N

hence, the tension in the string must be T =  Fc = 764.41 N

5 0
3 years ago
Need help asap
Vanyuwa [196]
I’ve done this before the answer is B
8 0
1 year ago
A person has a mass of 60 kg. What is the person’s weight in Newtons and in pounds?
liubo4ka [24]

Answer:

137.2 in pounds and in Newton's it's 588.399

3 0
3 years ago
An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

8 0
3 years ago
Other questions:
  • State the methods of nuclear fission waste disposal and the suitability of each method.
    12·1 answer
  • An electrically neutral balloon is rubbed on cloth and becomes positively charged. What can be said about its mass?
    10·2 answers
  • These waves are traveling at the same speed. Which wave has the highest frequency?
    6·2 answers
  • Two small insulating spheres with radius 3.50×10^−2m are separated by a large center-to-center distance of
    7·1 answer
  • Although 0 dB is often referred to as the lower threshold of human hearing, it is important to realize that the human ear is not
    10·1 answer
  • The graph shows the amplitude of a pausing wave over time in secondo (o).
    13·2 answers
  • How do you calculate the magnitude of the resultant force of two
    10·2 answers
  • What are the conditions of equilibrum
    14·2 answers
  • Elizabeth has two different electromagnets. Both electromagnets are connected to the same power source and are made of the same
    13·1 answer
  • The volume of the water in the graduated cylinder rose as some of the water was displaced by the table tennis ball. Find the vol
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!